What is the extreme f(x)=ln(y-x)+ln(y+x)+ln(16-y) ?

The extreme f(x)=ln(y-x)+ln(y+x)+ln(16-y) is Maximum(0, 32/3)

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extreme f(x)=ln(y-x)+ln(y+x)+ln(16-y)

extreme points

f (x) = ln (y - x) + ln (y + x) + ln (16 - y)

Maximum (0, \frac{32}{3})

Solution steps

Find the critical points:

(0, \frac{32}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{- 3 y ^{4} + 64 y ^{3} - 512 y ^{2} + 64 x ^{2} y - x ^{4} - 512 x ^{2}}{( y - x ) ^{2} ( y + x ) ^{2} ( 16 - y ) ^{2}}

D (x, y) = \frac{6 y ^{6} - 128 y ^{5} - 10 x ^{2} y ^{4} + 1024 y ^{4} + 256 x ^{2} y ^{3} + 2 x ^{4} y ^{2} - 2048 x ^{2} y ^{2} - 128 x ^{4} y + 2 x ^{6} + 1024 x ^{4}}{( - y + 16 ) ^{2} ( y - x ) ^{4} ( x + y ) ^{4}}

Check the sign of at each critical point

Check critical point

(0, \frac{32}{3}) :

Maximum

Maximum (0, \frac{32}{3})

f (x) = x^{2} - \frac{1}{2} y^{2} + 3 x

f (x) = (x - 3) e^{x}

f (x, y) = x ln (x + y)

f (x, y) = x^{2} + x y + y^{2} - 7 y + 16

f (x) = x^{2} (x - 2)^{2}

What is the extreme f(x)=ln(y-x)+ln(y+x)+ln(16-y) ?
The extreme f(x)=ln(y-x)+ln(y+x)+ln(16-y) is Maximum(0, 32/3)